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Why does elevator input move the turn coordinator ball in steep turns?
When practicing steep turns, I'm running into an unusual phenomenon.  If the airplane is losing altitude in coordinated flight and I apply aft stick pressure to pitch up and correct, I notice that the ball swings to the outside of the turn indicating a skidding condition.  Conversely if the airplane is gaining altitude and I release back pressure to the stick to descent, the ball always swings to the inside of the turn, indicating a slip.  This seems counter intuitive as one would think that an increase in AoA to gain altitude would result in increased adverse yaw, requiring more rudder pressure in the direction of the turn and less rudder pressure when the nose is lowered resulting in a lower AoA and adverse yaw.
I suspect the reason here has to do with the direction the nose is being forced in by the elevator input during the turn.  At a steep bank angle the nose is inadvertently being pulled inside the turning flightpath which results in a skidding condition, which should, therefore, be countered by less rudder pressure in the direction of the turn.  Conversely when elevator pressure is reduced, the nose would tend to stray outside of the tangential flightpath, resulting in a slipping condition.  The end byproduct of this would be reduced or even cross control inputs needed to maintain coordinated flight.  Can anyone else confirm this?

What kind of airplane is it?

C172, PA28, you name it.

Comments are not for extended discussion; this conversation has been moved to chat.

I'm assuming you are flying a single-engine prop plane with a nose-mounted engine turning in the "conventional" direction (clockwise as seen by pilot).  Is this true?  Have you demonstrated the phenomena to be consistently true while turning in both directions?

How steep?  60 degrees?  45?   I don't recall off the top of my head whether my own experiments spent much time above 45 degrees bank.  Possibly not.  (Maybe I can find some notes or video to check.)  Maybe that's the missing piece of the puzzle as to discrepancy in results.  Still not sure about the underlying "why" of it all though.

Also how transitory is the effect? In other words is it essentially only while you are adjusting the pitch attitude and airspeed? I.e. while the airspeed is *changing(? Or would you expect based on your observations to tend to see some skid whenever the aircraft is climbing even at constant airspeed and pitch attitude, and some slip whenever the aircraft is descending, even at constant airspeed and pitch attitude?

(Potential new ASE question-- are dynamics observed at, say, 30 degrees substantially different (other than in scale) than at 60 degrees?)

Actually MichaelH gave us all a good wakeup call on this question, rekindling memories of how fearful I was with steep turns cranking the ailerons way over and pulling so hard to hold the plane up, thinking surely inside rudder is going to send me down.  But the "eureka" moment came when it was realized, even in steep turns, power controls altitude.  In fact, being coordinated helps even more.

What I'm finding is that a pronounced forward movement of the stick/yoke that significantly drops the G-loading and makes the nose drop and the flight path arc downward does tend to move the ball to the inside quite markedly, but a pronounced aft movement of the stick/yoke that significantly raises the G-loading and makes the nose rise and the flight path arc upward has only a slight tendency to move the ball outwards, so slight that it's hard to be sure that it's not caused by something I'm accidentally doing the ailerons.  I don't have a good explanation for this asymmetry.

Observations were done in a glider, so it's not related to p-factor/torque/etc.  Gentle inputs such as you would do to arrest a slight loss or gain of altitude while circling in a power plane have such a slight affect that again, it's difficult to isolate possible affects of such on the ball, from something I'm accidentally doing with the ailerons by slightly changing the deflection.

Seems to me a related question could be asked without the "why" part, "does elevator input move the turn coordinator ball in steep turns?"  Soliciting corroborating evidence in a wide variety of aircraft.  As to the "why" part-- still not sure.  Footnote 3 in my answer?

(""move the ball to the inside"= move the ball toward the inside of the turn, i.e. toward the low wingtip)

In a steep turn you are making power changes, pitch changes and also making constant corrections with aileron to hold the bank angle, without even realizing it.  Once established in the turn some airplanes require a bit of in-turn aileron to hold the bank angle, some hold the bank with neutral aileron, and some require top aileron to keep from overbanking.  The top and down aileron inputs are inducing constantly changing adverse yaw forces which come and go, changing torque effects from the engine with power adjustments are producing changing yaw forces, and gyro precession from the prop from pitch motions are inducing yaw forces which come and go, and there are bumps, and you start running into your own wake.
In other words, you are in a machine pulled along by a big torque producing gyro that is slithering and sliding around in a gas with half a dozen forces and moments interacting on it simultaneously.  With all the subtle forces and inputs happening during a turn, I don't think you can identify and act on a single phenomenon like that.   What you have to do is, well, just do whatever it takes to hold the bank angle, altitude and center the ball and don't overthink it.
If you fly gliders, which use a yaw string that is more sensitive than a ball, even without the torque and gyro effects of an engine, the yaw string drifts this way and that while in a turn, seemingly independent of the aileron position some of the time. You generally make stabs of rudder in concert with aileron, but sometimes the yaw string seems to have a mind of its own and you just do what you have to do with your feet.

It is important to realize the "ball" is simply rolling back and forth in a curved glass tube to
indicate the direction of net G forces.  It is also known as an inclinometer.  What is happening in your steep turn is the elevator/wing orientation is now at an angle to gravity, so pulling "up" also tightens your turn radius, forcing the "ball" to the outside.  In a steep turn the  elevator becomes more "rudder-like".  If you roll to 90 degrees, the elevator IS your rudder, and the rudder will pitch your nose up or down.
Your thought to "step on the ball" is correct, as rudder input will  bring the nose back into the line of coordinated flight,  helping hold altitude.  Try adding a bit more power too.
Pulling alot more elevator to hold altitude in a steep turn is not good technique as it can lead to a stall or spiral dive.  Sometimes just rolling to a slightly lower bank angle will do the trick.
I would review this with an instructor, but a little "rudder to the sky" may help here.  Get coordinated and see how much throttle you need to stay level.

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If the airplane is losing altitude in coordinated flight and I apply
  aft stick pressure to pitch up and correct...Conversely if the
  airplane is gaining altitude and I release back pressure to the stick
  to descent

What you really want to do is to change your technique.
Your goal (assuming for example a steady 60 degree AOB) is to roll into the turn, set your G (2G = what you need for 60 degree AOB) and then lock in the back pressure while varying your AOB to set your nose position relative to the horizon. Climbing? Over bank a little to let your nose slice down, and then reset the bank so it holds fixed to the horizon. See what that gives you. Descending? Do the opposite. But don't pump the nose to go up or down. Particularly if you are near a stall AOA, it can cross you over and you can lose control.
This will smooth out your flying and result in constant radius turns.

Absolutely, no rule says it has to be exactly 60 degrees.  Much safer.

Also, power can be added or taken out to control altitude.

I had a long answer on why the ball went one way or another and posted it, and then realized I was completely wrong. :) Still puzzling on it.

When practicing steep turns, I'm running into an unusual phenomenon.
If the airplane is losing altitude in coordinated flight and I apply
aft stick pressure to pitch up and correct, I notice that the ball
swings to the outside of the turn indicating a skidding condition.
Conversely if the airplane is gaining altitude and I release back
pressure to the stick to descent, the ball always swings to the inside
of the turn, indicating a slip.

This question surprised me.  Your observations are in fact consistent with a statement by Wolfgang Langewiesche in his aviation classic "Stick and Rudder".  He claimed that while turning1, a sideslip is often caused by "insufficient lift" and should often be fixed by increasing elevator back pressure rather than adding inside rudder.  But I've never found his claim to be consistently true and I've long believed that it is one of the few significant errors in his otherwise excellent book.2
I have found that very pronounced elevator inputs while turning that create very significant changes in G-loading do produce brief, small excursions of the slip-skid ball in the direction predicted by Langewiesche's idea.  I believe these are produced by complex three-dimensional changes in the flight path that are beyond the scope of simple analyses.  Or to take a somewhat related situation, in a wingover involving a full 90 degrees of bank, if the aircraft is allowed to "float" over the top at a rather low G-loading (involving little or no aft pressure on the control stick), with no bottom rudder applied, the slip/skid ball and or yaw string will indeed show quite a pronounced slip as the aircraft "falls" earthward.3
But in simple turning maneuvers involving only moderate changes in the direction of the flight path in the vertical plane (i.e. only moderate changes in climb or descent rate), I've not found elevator inputs to play a role in turn coordination (in the sense of avoiding slips and skids) in any consistent manner.  (In single-engine prop planes, there definitely can be an effect due to p-factor, but the direction of the effect will depend on the direction of turn.)  I've done rather extensive tests to look at this in several single-engine airplanes and ultralights (including some with a high-mounted pusher engine), and several different sailplanes.
The topic has long been of special interest to me, because in the hang gliding community, at least in the US, there's long been an idea that the "correct" pitch inputs while entering a turn will keep the flight path "coordinated", while inadequate pitch inputs will cause the glider to "sideslip".  Along the same lines, an intentional nose-down pitch input while the glider is entering a turn, or is turning, is thought to produce a very pronounced "sideslip". (If true, this would be compatible with Langewiesche's idea as noted above.)  Many pages have been devoted in training manuals and magazine articles to "explanations" of these supposed phenomena.  While it makes some sense to use the word "coordinated" to describe the harmonized use of all available flight controls to produce the desired results-- and the only available control inputs in hang gliding are weight-shift pitch and roll inputs-- I've found the idea that pitch inputs have something to do with causing or preventing sideslip in hang gliders to be a complete "red herring".  Some hang glider pilots appear to be mistaking the sensations of a diving, accelerating turn for the sensations of a true "slipping" turn.  Extensive experiments with yaw strings and slip-skid balls on hang gliders suggest that sideslip in hang gliders is related almost entirely to roll rate, with a slight residual sideslip remaining when the bank angle is constant-- and this slight slip seen in a constant-banked turn tends to be greater when the airspeed (and sink rate) are low than when the airspeed (and sink rate) are high-- exactly opposite to what the community "conventional wisdom" (and Langewiesche's idea) would predict.  Part of the confusion may be due to the fact that in some hang gliders, especially those with pronounced anhedral, "pulling in" the control bar while rolling from wings-level into a turn can produced a marked increase in roll rate, thus indeed producing a rather pronounced sideslip, which ends very soon after the roll rate is brought back to zero.  Anyway, the long and short of it is that sideslip in hang gliders is produced primarily by adverse yaw while rolling, and for a given roll rate, this slip due to adverse yaw is not reduced by moving the control bar forward to "coordinate" the turn and reduce the gain in airspeed and loss of altitude.  Pulling the bar aft while holding the glider in a constant bank doesn't make the glider sideslip as it dives and accelerates.  (End of tangent-- but may help the reader understand why I've spent quite a bit of time exploring these relationships in more "conventional" aircraft as well.  And the basic results were the same in all the aircraft tested.)
I'm assuming you are flying a single-engine prop plane with a nose-mounted engine turning in the "conventional" direction (clockwise as seen by pilot).  Is this true?  Have you demonstrated the phenomena to be consistently true while turning in both directions?
At the risk of contradicting both the original OP of the question and the esteemed Wolfgang Langewiesche, I'd suggest that this is not an effect that can be consistently replicated in both directions in a wide variety of aircraft.  Would be most interested to be directed to video footage or other verification that suggests otherwise-- ideally with pilots feet clearly visible off the rudder pedals!

Can anyone else confirm this?

I'm not sure whether the OP meant (by "this") his theory was to what was happening, or simply the observed phenomena, but I'd sure like to see answers posted (or chat replies) by other ASE members who are able (or unable) to confirm the observed phenomena-- with specifics as to bank angle, and whether the effect is transitory (connected to changes elevator back pressure and airspeed) or lasting (enduring as long as aircraft is climbing or descending, even with pitch attitude constant), and confirmation that the effect was observed in both directions of turn, etc-- regardless of what the underlying mechanism is believed to be.4
Footnotes:

Langewiesche didn't specify that the turn must be steep in order for this phenomena to be observed.  OP mentions "steep" turns.  It's possible that my own experiments described in this answer were mainly focused around turns of 45 degrees bank or less-- need to check notes for more details to refresh memory.

Similar content-- perhaps inspired by the content in "Stick and Rudder"?-- appeared in some (circa 1969?) editions of a book called "Modern Airmanship", edited by Neil D. Van Sickle.  In later editions, such as the 1999 8th edition edited by John F. Welch, Lewis Bjork, and Linda Bjork, this content has been removed.

We can think of slips as fundamentally being caused by either of two phenomena: (one), an aerodynamic asymmetry creating a yaw torque that is only brought into balance when the aircraft is flying in a slipping condition, with the nose yawed to point to the "outside" or "high side" of the actual direction of the flight path, or (two) a sudden demand for an increased yaw rotation rate (if the nose is to remain aligned with the flight path), which the aircraft's yaw rotational inertia tends to resist, until a large enough slip angle is achieved to create a significant yaw torque to overcome the yaw rotational inertia and increase the yaw rotation rate.  As the airspeed reaches a minimum at the top of a wingover, and aerodynamic forces thus are low, while the rate of earthward curvature of the flight path (due to gravity) reaches a maximum, the latter effect probably best describes the reason for the sideslip seen at this point.  The steeper the bank angle, the more the earthward curvature of the flight path demands a yaw rotation rather than a pitch rotation, if the nose is to remain aligned with the flight path.

A few more words about the possible "underlying mechanism", or lack thereof, for increased back pressure to tend to cause a skid, and decreased back pressure to tend to cause a slip, as OP has reported --  the immediate effect of increased back pressure on the stick or yoke would seem to be an increased aerodynamic force that acts "straight up" in the reference frame of the inclinometer (slip-skid ball) tube, and so should not influence the deflection of the slip-skid ball. (Keep in mind that the slip-skid ball does not respond to some sort of "balance" between aerodynamic forces and gravity-- rather it responds to aerodynamic forces alone.)  The longer term effect of increased back pressure on the stick or yoke is a lower airspeed, and thus (for same bank angle) a lower turn radius, a higher turn rate, and a higher required yaw rotation rate.  It would seem that when extra back pressure is applied on the stick or yoke and the airspeed drops and the radius of curvature of the flight path decreases, and the turn rate increases, and the required yaw rotation rate also increases, the aircraft's yaw rotational inertia would have some tendency to keep the nose yawed too far toward the "outside" or "high side" of the flight path (basically a "lagging" effect) which should produce some temporary sideslip (not skid). Yet OP's observation is that increased back pressure tends to cause a skid. I'm at a loss to reconcile this  discrepancy.  Of course, the steeper the bank angle, the less changes in turn rate involve changes in yaw rotation rotation rate at all-- at steep bank angles the dynamics at play mainly involve the pitch axis.

Would love to see this topic taken further.  Maybe we can set up a chat room to post video links to experimental results?

Imagine a banked plane in four scenarios. In each scenario the plane maintains the same vertical speed (it does not accelerate up or down).

The plane is banked but not turning. The aircraft creates 1G of lift by angling through the sky in an uncoordinated manner and does not turn at all. The direction of lift is pointed straight up, and the ball moves to a position opposite it, pointed strait to the ground.
The plane is banked, but in a slip. The aircraft does turn but not as much as it should given the angel of bank. The aircraft is generating 1G of lift in the vertical direction, but not enough lift in the horizontal to put the combined vector perpendicular to the wings. The ball falls partway towards the ground.
The plane is in a coordinated turn. The aircraft is generating 1G of lift in the vertical direction and just enough lift in the horizontal direction to put the combined vector perpendicular to the wings. The ball is centred in the gauge.
The plane is in a skidding turn, it is turning faster than it should given the angle of bank. The aircraft is still generating 1G of lift in the vertical direction, but now it is generating more than the required amount in the horizontal direction, and the combined vector is past perpendicular. The ball moves to the outside of the turn.

The essential part here is that unless the aircraft is entering a climb or coming out of a dive, it is producing 1G of lift in the vertical direction. If the plane increases or decreases the amount of total lift produced, the direction of the vector will change accordingly. If the direction of the vector does not match the angle of bank you will be uncoordinated. Too much bank, or not enough lift gives a slip, too little bank or too much lift gives a skid.


text:
Can all airliners be turned without rudder input?
Is it possible for pilots to make a right or left turn, using just the ailerons and without rudder input in all airliners? If not why so, for what reason?

Your wording of this question "rudder input" instead of "rudder movement" makes it tricky or interesting to answer because this is actually a feature of advanced fly-by-wire system.

@user3528438 It is not a feature of an "advanced" system. Turns on any yaw-damper equipped airliner (707, DC-8 to today) should be flown without any rudder input. I think this is more along the lines of the answer that is being sought, rather than the complex aerodynamics and glider answers so far.

@jjack "roll-yaw decoupling"

It depends on the aspect ratio of the wing and the lift coefficient. Short, stubby wings at low angle of attack will not create much adverse yaw with aileron input, especially when the ailerons have differential linkage (more deflection on the trailing-edge-up side and less on the opposite side). Differential gearing is also beneficial in reducing stick forces, so many small aircraft with manual controls employ it. Therefore, turning with ailerons alone is possible with low aspect ratio wings.
Airliner wings have aspect ratios between 7 and 11, so here it is helpful to reduce the lift coefficient by flying fast and lower than at maximum cruise altitude if you want to turn with the ailerons alone. If you allow rudder movements but just want the pilot to rest his/her feet on the floor: Almost all jets have yaw dampers, that is a device which will deflect the rudder automatically such that yaw is minimized. Therefore, airliners can be flown with the feet off the pedals and still ailerons and rudder will keep the plane in a coordinated turn.
With gliders and their high aspect ratio wings, however, using only ailerons will mostly yaw the aircraft, especially at low speed when the lift coefficient is high. The wing also has high roll damping from its high aspect ratio and low flight altitude, so the roll angle achievable before yaw becomes dominant is small. The yawing condition will now create its own rolling moment due to the dihedral effect of the wing. Unfortunately, this effect counters the intended rolling motion and prevents the aircraft from turning with ailerons alone. If adverse yaw is large enough, the aircraft will not even begin to roll but will only yaw with aileron input.  Some glider pilots even apply cross controls when entering a thermal in order to maximize their rolling speed (rudder into the turn and aileron against to provoke more yaw and let the dihedral effect roll the glider).
If you find yourself in a high-performance glider with a jammed rudder, speed up in order to fly turns with the ailerons (and elevator) only. The lower lift coefficient will reduce adverse yaw so direction can be controlled with ailerons and elevator only.

I have just noticed that the question text (but not title) mentions airliners. From riding in the back of many airliners, I have noticed that inboard ailerons are used in cruise,  and spoilers may be used in conjunction with ailerons, at least when the flaps are extended. I would guess that these would both reduce the adverse yaw? And perhaps also so would the yaw damper?

@sdenham: Inboard ailerons are a specialty of Boeing and their main purpose is to reduce wing torsion. A high true airspeed reduces roll damping, so very little rolling moment is needed and adverse yaw is low at high speed. Spoilers are used mostly in the approach phase and open the gap between wing and flaps. In that case roll control with spoilers does indeed lower adverse yaw. A yaw damper works on the rudder so the pilot doesn't have to, so this would be against the spirit of the question.

@jjack: In a way yes, inboard ailerons cover the kink in the trailing edge where inboard and outboard flaps would collide. They move trailing edge down when flaps are extended and as ailerons at high speed. True flaperons, however, mix aileron and flap inputs. The inboard aileron is only one of each at a given time.

+1: Nice answer. When you write 'prevents ... from turning...', do you mean prevents a sustained turn? I would have thought that a series of rolling & restoring actions would result in a (very slow) sustained turn?

@copper.hat: Once the aircraft has rolled, turning is inevitable. Adverse yaw can indeed prevent the aircraft from rolling such that no turn results, not even a very slow one. Maybe I should clarify that part.

Cross control in gliders while doing tight circles (usually while in thermals) is required in order achieve a coordinated turn. The inside wing's air speed is lesser enough than the outside wings air speed that opposite aileron is required to prevent the glider from banking further inwards. Rudder input is used to achieve a coordinated turn. The simplest indicator of a coordinated turn is a streamer that flows up the windshield: in a coordinated turn, the streamer will go straight up the windshield from the pilot's perspective, indicating that the glider is not in a skid or a slip.

@rcgldr: I know, but that is not what I meant. Read the answer again.

@PeterKämpf - for a normal turn in a glider, it would seem that you'd want a coordinated turn, so just enough rudder to keep any yaw related effects from resulting in a skid or a slip. I'm not aware of outwards aileron input used during a roll or turn except to reduce or stop a roll due to the relatively lower air speed on the inner wing.

@rcgldr: This is about a short jerk on the stick against the desired roll direction when entering the turn. This jerk supports the rudder (which is too small at low speed in almost all gliders) in creating the desired yaw. Once dihedral rolls the plane, you support it with the stick in the right direction so roll damping will not stop the rolling motion. The effectiveness of this manoeuvre is disputable, but I found enough pilots who were sure it helped them to enter a thermal more quickly.

@rcgldr: When in a stable turn, you want the string on the canopy to point slightly outwards since its location is ahead of the cg, so it sees some sideslip when the whole glider flies a coordinated turn.

@PeterKämpf - true about the yaw string in a turn, slightly outwards as you commented. I've also read there's a trade off between outwards ailerons versus a small amount of side slip to use dihedral effect to act as outward aileron input to prevent a bank from increasing due to the relatively lower air speed of the inner wing.

@rcgldr: Yes, that is correct, even though I never found a good proof that this sideslip is better than flying straight. Also, the mass distribution along the span can help to reduce aileron input.

@PeterKampf - I've had an experience in Challenger ultralight, flying feet off rudder pedals, where was banked left maybe to 10 to 15 degrees, holding full left aileron to try to increase bank, adverse yaw was so bad that the yaw string was streaming way out to right,  aircraft was in slow right turn.  Could also be replicated in other direction.
Only with doors on, no doubt due to increased forward side area of fuselage, and resulting decreased yaw stability.  Similar in Schleicher  Ka-6 sailpanes at some airspeeds- the model with the higher forward fuselage.  (Cr as opposed to E?)

@PeterKämpf - above was re "Once the aircraft has rolled, turning is inevitable."

For most aircraft, it is. Deflecting ailerons only results in adverse yaw: the aileron deflecting downwards has higher drag than the one deflecting upwards, and the nose turns away from the direction the pilot wishes to bank into.
Image source
The relative magnitude of this effect depends on wing span and airspeed: hardly noticeable in jet fighters, highly disturbing in gliders.

Adverse Yaw does not cause aircraft turn. It only causes yaw. Yaw is not turn. Yaw is a deflection of the aircraft flight path from the aircraft longitudinal axis about the vertical axis of the aircraft. Turn is a continuous change in the aircraft light path direction of motion through the air mass.  The aircraft can be in a continuous established sideslip angle as a result of a yaw inducing force, and still be maintaining a constant heading... i.e., zero rate of turn.  Lift, pointed in a direction other than perpendicular to the horizon, is what creates turn. Nothing else.

@CharlesBretana you'll need to tilt your lift vector to compensate for rudder input if you want to fly in a straight sideslip path: bank to left, right rudder pedal. If you don't bank to left and give only right rudder pedal, this causes an angle of sideslip which will turn the plane as long as you press on the pedal.

@Koyovis. This is only true if the force from the rudder is different from the sideways force from the sideslip angle. With no other yaw force on the aircraft, if you have right rudder, that does create a leftwards force, but it also creates a right yaw (sideslip) which itself creates a rightwards force that opposes the leftwards force from the rudder. In a multi-engine aircraft with one engine inop, you need bank IN the direction of the rudder to compensate because the Yaw from the rudder just eliminates yaw from operative engine, and sideslip is zero.

@Koyovis, It works in exactly the same way as it does in pitch. The elevator pulls the tail down and, because it's behind the CG, creates a nose up pitching moment (torque). Lift on the wings lifts the aircraft up, and because it's also (generally) behind the CG it produces a Nose down pitching moment. Because the moment arm of the elevator is so much longer the torques balance, but the Up force from wings is much larger than the down force from the elevator and the aircraft is able to fly. The force from the rudder, and the force from the sideslip angle of the fuselage act in the same way.

@Koyovis, When you say "Deflecting ailerons only results in adverse yaw: .... , and the nose turns away from the direction the pilot wishes to bank into.", you are conflating Turn, and Yaw. The nose of the aircraft does not "turn" away from the "direction the pilot ... " it only "yaws" away, and it only yaws away until the sideslip angle reaches the point where equilibrium occurs.

@CharlesBretana I'm sorry, I'm out of this. I can see that there is no doubt in your mind that you are right. Instead of teaching me how you think that aerodynamics 101 works, why don't you verify what you say. Google it for instance, or ask a flight instructor. The rudder alone can turn the flight path of the aircraft, pilots don't do it because of the unpleasant side forces like turning a car. I won't be participating in any further discussion.

@Koyovis, Good day sir.

@CharlesBretana feel free to add an answer to this question

In fighter simulators I almost never used the rudder: I just rolled to almost 90° then applied the elevators...
IRL I tried this with gliders (we even had to learn how to fly if one of the 3 controls were lost - elevators are the worst, one of the other two you can get by without much easier). You can turn with gliders really well by just using the ailerons, and compensating with the elevators to not drop your nose too much (and if you are rolling, then your elevators are partially working as rudders anyway). Not good enough to do thermals with great efficiency, but good enough to get back home and land.

In fighter simulators I almost never used the rudder: I just *yawed* to almost 90° then applied the elevators - rolled, surely...?

@GrimmTheOpiner : yes, I overlooked it, English is not my native language

In the F-4 Simulator, (before leading edge slats were added), the adverse yaw was so extreme at high Angle of Attack that use of aileron would put the aircraft out of control. To roll the aircraft at high Angle of Attack, we used the rudder exclusively. As an instructor, in fact, when a student was flying the aircraft at high AOA, (like for stall series maneuvers), we would mash our knees together on both sides of the stick to ensure that the student did not introduce any aileron.

Very short answer: Yes, you can turn an airplane without using rudder input. Simply rolling the airplane to an appropriate bank angle, combined with applying aft stick pressure to maintain altitude, will cause an airplane to turn.  The horizontal component of the lift vector the wings banked in the direction of the turn will cause the airplane to turn in the direction of the bank.

(source: flightliteracy.com)
As to maintaining a coordinated turn, that’s another story altogether.  Rolling an aircraft using ailerons will generate an imbalance in induced drag between the two wings, causing adverse yaw.  Other solutions can mitigate this, such as differential spoilers, as in the case of the MU2 aircraft.  Airliners also use this method for lateral control during high subsonic flight encountered in cruise.  As Peter Kämpf pointed out airliners also use yaw dampers to mitigate other undesirable flight characteristics in high speed cruise so they can and are maneuvered using no rudder pedal input from the pilot.  In propeller driven airplanes, the left or right turning tendencies caused by the propeller will almost always result in an uncoordinated turn if no rudder input is used.  But the airplane will still turn in the direction it’s banked regardless.

Lots and lots of radio-controlled model airplanes lack rudders and fly just fine.  Some are even shaped like airliners.  No reason this shouldn't scale up to full scale.  (The point being that this strongly suggests that airliners can turn without deflecting the rudder, not that there's "no reason" not to design a full-scale airliner entirely lacking a rudder.)

@ymb1 - edited to address your comment

One anecdote, if I may. I mostly flew trainer R/C planes with full span ailerons. These had little adverse yaw (not very noticeable from the ground) and I was not accustomed to coordinating turns. But one day, a friend let me fly his Carl Goldberg Piper Cub. The prop came off on climb out, and I had an awful time getting it to turn back to the field. The adverse yaw was so bad it almost wouldn't turn with aileron alone. I managed to get it back and land it, though.

@FredLarson -- had similar experience w/ HobbyZone Super Cub I modified by adding ailerons -- not full-span-- plane was originally designed to fly rudder-only and hence had very marked dihedral.  Usually folks remove the dihedral when mod by adding ailerons, I didn't.  Well, it did a good job of training me to use rudder along w/ ailerons!  But then w/ most other a/c, I found there was no need to keep doing that.

I just came across your question by accident regarding turning an aircraft with the use of rudder.
The simple answer is YES.
You NEVER lead a turn using Rudder. Its a combination of of Aileron to lead into a bank (left or Right). Back pressure is then applied on the yoke/ stick to maintain altitude. Reason, additional drag encountered resulting in a loss of lift. Back pressure provides this additional lift required to maintain altitude.
The rudder's part is all this is to balance the turn. This means that aircraft "slip" or "skid" during a turn. Basically slip into towards the centre of the turn or skid outwards not maintaining a fixed radius turn in otherwords.
The "Ball" on the Turn Coordinator should be centered during the turn. So if the ball go to the left "then stand on the Ball" = left rudder to balance and coordinate your turn.
All this is a very smooth coordinated movement in practice.
If I may, just go one stage further regarding Adverse Yaw.
I also see some comments on the subject. Adverse yaw is one of these "Strange but True" issues.
Firstly, if you were to gently deflect the rudder (the Action) either side of center in normal controlled flight, two things will happen:

It will produce a movement about the vertical axis.This is the Yaw.
The Primary effect of rudder.
Lots of groaning and moaning from the aircraft as if to say, Hey what are you doing to me ?.. reason the aircraft is being knocked out of balance.

The Secondary effect of rudder is to Roll. If you prolong the defection of the rudder the aircraft initially yaws followed swiftly by a rolling movement about the longitudinal axis.. dangerous if maintained..spiral dive follows..
Now, the main part. The adverse bit.
When a turn is initiated using aileron to lead into the turn the up-going aileron enters a reduced pressure area. Remember relatively low pressure on top surface of wing. While the down-going aileron enters relatively higher pressure area.
For example: Left turn.
The left aileron is deflected upward and the right aileron moves downward.
As the chord line on the left wing is altered in an upward direction
(The Chord line runs from the leading edge directly to the trailing edge of the wing or a control surface.), this effectively reduces lift on that wing. The opposite happens on right wing.. Lift increases. Result a left-hand bank should occur.
But, oops. a Surprise awaits us, We started a Left turn using Aileron but we initially get a RIGHT YAW effect..just as if we applied Right Rudder. So now after scratching our heads for a while, lets see what happened there....
In brief, as the lift changes on the outer parts of the wing, so also does the associated drag. Known as Induced Drag. A By-product of lift.
So the down going left wing, has reduced lift and reduced drag by association
Right wing has increased lift and increased Induced drag.
Net result drag on left side - Reduced. Right side increased, so its acts like a "brake effect" on the right side, and relieved on the left side. Produces a Yaw to the RIGHT initially.
Solution: In aircraft nowadays the angular movement of the ailerons is arranged so that the up-going aileron travels further up than the down-going one. For example full up deflection say 50 degrees, Full Down movement 25 degrees and proportional in between.
This limits the effect of Adverse yaw due to up-going aileron traveling further into lower pressure area and likewise, down going moves down less in higher pressure area as the Chord line is altered differentially. This system is known as Differential Ailerons.
One further development on this was the Frise Differential Aileron, which allows the top side leading edge, of the aileron, to protrude into the airflow on the down-going aileron, creating a Parasite Drag on that side to counter the Induced Induced drag on the other wing. Combined its known as the Frise Differential Aileron.
You might need to read the last paragraph again, but the penny will soon drop for you.

Welcome Gerry: That's a good description of adverse yaw, but not the topic of the question (turn without rudder). You may need to re-focus your post.

Thank you. I put in an brief article a few hours earlier on the main topic regarding turning using rudder. This was just an add on as i noticed alot of comment on adverse yaw. I did prefix this by saying i was going one stage further prior to adding this bit. Gerry

Correct indeed, but that's not how SE works, we prefer only answers that are directly answering the question, hence "comments", even good ones, may be deleted.

Gerry, the Frise aileron is about lowering control forces. The parasitic drag you describe actually mostly occurs outside of the normal operation range of a Frise aileron.

I've some small experience on the MU-2 series aircraft, which has no ailerons. It uses full span flaps and spoilers. The autopilot can reflex flaps to control roll and function as a wing leveler, but is not available in landing configuration.
Moreover, there are large amounts of mass at the outer wing, considering that there are wing tip tanks. This short-coupled twin will give you an appreciation  for controlling the direction of the total lift vector and using all available runway on takeoff.
Using rudder to control small rolling moments when close to Vs is the second thing I learned training in simulator; the first is to stay away from Vs. Being close to Vs, stalling the rising wing by using spoiler roll control results in a snap roll into the rising wing creating a stall condition on one side and once there you become another stall/spin statistic. It is difficult but not impossible to reestablish the lift vector to the centerline of the aircraft – the first order of business – given time, space, and engine power in order to recover. This is usually not available in landing configuration.
During a turn, the yaw damper and down wing spoiler will coordinate the turn with feet on the floor. You must use nose up elevator to compensate changes in lift vector caused by a sustained maneuver. I have noticed in flight the almost zero displacement of the rudder during a two-minute turn at cruise airspeed. That was some time ago, but my experience, for what it's worth.

Please edit this answer so that it's clear what your answer to the stated question is. Following that up with a narrative as to how you arrive at that answer is fine, but narrative alone is hard to follow. Paragraph breaks and proper sentence capitalization would be helpful as well.

Of course! Neither the rudder, not the ailerons, actually turn the aircraft. It is LIFT that turns the aircraft. It "turns" the aircraft in the direction that the Lift vector (perpendicular to the wings) is pointed in.
So to turn, you need to point the Lift vector in the direction you wish to turn, (roll into a bank) and apply some Lift (Pull back on the stick/yoke).
To turn up (increase pitch attitude) leave the aircraft in zero bank angle, and pull on the yoke. To turn left , place the aircraft in a left bank, and pull on the yoke, etc.
To roll into a bank in a coordinated manner, you generally need both ailerons and rudder, but if you're not picky about being uncoordinated,  you can roll into a bank with either one.

The airstream can exert forces and moments in 6 degrees of freedom on the airframe, lift is only one of them.

Any forces exerted by the Airstream on the aircraft are Lift.  And they can be exerted in any direction. The rudder produces both Lift and Yaw, but these are not the same. None of these Lift forces are yaw. Yaw is a measure of the deflection of the fuselage from the flight path about the vertical axis of the aircraft. Yaw is a torque - a force causing a rotation, Lift is an ordinary force, causing an acceleration.  Defining Lift as only that component of all aerodynamic forces perpendicular to the wings is an artificial engineering simplification to aid in visualization and computation.

Aircraft can be turned by the rudder alone: it creates a torque about the z-axis of the aeroplane. Not sure where you got your references from when you say Any forces exerted by the Airstream on the aircraft are Lift.

@jjack, Actually, I would.  The "force" we call drag is a fiction, it's just another artificial construct invented by engineers  and Mathematicians to make it easier to think about lift and perform calculations. All Aerodynamic forces are applied locally, to each point on the aircraft. Lift itself is just a artificial summation of all the tiny aero forces applied all over the surface of the aircraft, and "drag" is just the mathematical component of all those tiny forces that lies parallel to the flight path. ("Lift" is the component perpendicular to the flight path).

@Koyovis. torque is NOT the same thing as Force. Force produces acceleration, and can effect a turn (rate of change in heading) torque (without a total non-zero force) cannot - it can only change attitude or orientation. Yaw from the rudder produces a torque - which changes the sideslip angle until the sideslip angle has increased to the point where the fuselage has generated enough opposite torque to create equilibrium. If, at the point where torque balances, the contribution of force from the rudder is equal to the contribution of opposite force from the sideslip angle, no turn results.

@Koyovis. True. And I see that you are fully aware of these factors. But I am not nitpicking, or trying to obfuscate something of significance. There are indeed, two critical aspects to this issue. Force, and Torque, and they are not the same thing. and an argument that conflates them has the potential to be misleading at best, and at worst, wrong.  To understand the torque, I have to think about where on the aircraft the forces (from the rudder, or from the fuselage sideslip) are being applied.

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text:
During inverted banking, what happens to heading?
In normal flight, a plane that banks will turn and change heading.
What happens if the plane is inverted?
If the plane rolled 180° to wings-level inverted flight, then it will not be turning.
But if the plane rolled 135°-- 45 degrees beyond wings-level inverted flight-- what would happen?
Will the aircraft's heading tend to change toward the earthward wingtip, or toward the skyward wingtip?
Does it matter if the wing is positively loaded (e.g. the aircraft is performing a barrel roll) or negatively loaded (e.g. during a "slow" roll, or while performing a banked turn during sustained inverted flight?)
During sustained inverted flight, would the pilot be pushing the stick forward to maintain altitude?

Are you assuming the aircraft is maintaining altitude? The answer to this question will depend on the direction of the lift vector.

This question needs fixing. You've stated what the bank angle/ roll attitude (same thing) is.  What then is the meaning of the statement "And thus bank in the direction the wheels are pointing?"  Are you asking which direction the nose will swing, i.e. will nose track around the horizon such that the heading changes in the direction that the wheels are pointing, or the opposite direction?  That is not the same as asking in what direction the a/c will "bank".  I have video fresh of an rc plane taken 2 hours ago that answers the former question.  Is that your question then?  Please clarify.

Anyway as far as heading changes go, assuming the planes is negatively loaded-- e.g. in stabilized constant-airspeed inverted flight-- the heading will change toward the earth-ward wingtip, which as seen from the cockpit, looks like a turn in the "backwards" direction as normal.  For example if you are wings-level (inverted) and the move the stick or yoke to roll into a turn, the heading will start swinging in the direction OPPOSITE to the direction in which you moved the stick or twisted the yoke.  But, the question does need fixing as noted above.

If you can fix the question I'll try to transform these comments into an actual answer.

Sorry "needs fixing" was a little abrupt but the question can be improved by deleting or clarifying the question about will the plane "bank the way the wheels are pointing".  You seem to be asking which way the flight path will curve-- you've already stated what the bank angle is.

Oh I see this is very old question.  I will edit it myself.

We'll see if that is approved or not--

Would a plane in 135° roll be using stick-forward to retain altitude? And thus bank in the direction the wheels are pointing?

Yes and yes.
Let's say you're flying North.  You roll inverted, then lift your left wing 45 degrees above the horizon (this is the wing that pointing East, now that you're upside down).
You'll experience the usual G forces pushing into your head (it hurts), and the plane will start turning to the West.
I once saw a great diagram that explained which controls do what when inverted - it showed the outline of the plane with one cartoon pilot sitting in the normal position, and another "phantom" pilot sitting in the same seat with feet on the rudder pedals, holding an extension of the stick, but with his head pointed the opposite direction - head towards wheels instead of away from wheels.
When the plane's inverted, the "phantom pilot" is upright, and all the controls would behave normally from his point of view.

You say to lift your left wing, the wing that was pointing east. I think you mean your left wing, the wing that IS pointing east now that you're inverted (or that WAS pointing west prior to rolling inverted).

@ammPilot - You are correct.  I've updated the answer - thanks for the clarification.

I'm guessing the phantom pilot was facing the same way as the real pilot (towards the nose of the plane), right?

@TannerSwett Yes

Concise answer, but why do you repeat the statement from the question "And thus bank in the direction the wheels are pointing?", and then give a "yes" answer?  I don't even know what that means.  It seems to me to be a reference to rolling or changing bank angle.  Roll response has nothing to do with direction of G-load.  Hopefully the question will be improved.

A concise way to describe the situation is that during sustained inverted flight, if the aircraft is banked away from wings-level inverted, the flight path will curve toward the wingtip that is closer to the earth.  This is a bit odd because the flight path ends up curving AWAY from the direction of the initial roll, i.e. away from the  direction that the pilot moved his control stick or twisted his control yoke to start the roll away from wings-level inverted.  During inverted flight, if we bank aircraft toward the left, from our perspective sitting in the aircraft, the nose will start tracking around the horizon toward the right-- toward the earthward wingtip.
An interesting consequence of this is that during sustained inverted flight, to neutralize adverse yaw and keep the nose lined up with the flight path as we roll (change the bank angle), we have to apply rudder OPPOSITE to the direction that we are moving the control stick or twisting the control yoke.
However, lest we lead the reader astray, let's note that the adverse yaw issue in practice isn't so severe as to prevent the aircraft's heading from swinging in the direction of the earthward wingtip, perhaps after just a slight bit of hesitation.  In other words as we fly wings-level inverted and we move the control stick to our left, we might see the nose swing a few degrees to our left for just an second or two but then we'll see it begin to track around the horizon to our right, toward the earthward wingtip, as the flight path curves toward the earthward wingtip, even if the nose remains yawed slightly to the left of the actual direction of the flight path at any given instant, until such time as we end the rolling motion and stabilize at a constant bank angle, at which point the adverse yaw effect largely vanishes and the nose tends to become (almost) fully aligned with the actual direction of the flight path at any given instant as the turn continues.  The whole adverse yaw issue is much more likely to be noticeable in a long-winged aircraft like an aerobatic sailplane, than in something like a Pitts Special.  (In the Pitts the required rudder inputs will probably have more to do with engine torque and P-factor than with aerodynamic adverse yaw!)
And yes, this is all a result of the fact that we've put the wing at a negative-lift angle-of-attack, usually by pushing the control stick or yoke forward.  If we unload the wing to the zero-lift angle-of-attack (at which point we'll go weightless in the cockpit), adverse yaw vanishes and we don't need to coordinate our roll inputs with rudder inputs.  We also no longer see the flight path curve toward the earthward wingtip.  If the wing is at a positive-lift angle-of-attack during inverted flight (e.g. at the top of a positively-loaded barrel roll), to keep the nose lined up with the flight path, we'll apply rudder in the same direction as we're applying ailerons.  We'll also see the flight path curve toward the skyward wingtip rather than the earthward wingtip.
The easiest way to quickly understand all this is to "fly" through the maneuver with a little hand-held model of an airplane, thinking about which way the pilot is moving the control stick, which direction the bank angle is changing, and which way the wing's lift vector is pointing.  Whichever direction the wing's lift vector is pointing, is the direction that the flight path will tend to curve, causing the nose to tend to track around the horizon in that direction.  And as far as adverse yaw goes, and the direction of required rudder inputs, keep in mind that as the aircraft rolls, the wing that is moving (rising or falling) AWAY from the direction that the lift vector is pointing tends to "drag back"-- actually this has to do with "twisted lift" (see https://www.av8n.com/how/htm/yaw.html#sec-adverse-yaw ) as much as actual drag-- requiring a rudder input toward the other wingtip.  Again, fly it through with a hand-held model and it will become clear.
In relation to adverse yaw and required rudder inputs, this answer could reference "keeping the ball centered", but note that during negatively-loaded flight, the tube of a conventional slip-skid ball curves the wrong way, so during sustained inverted flight, the ball will tend to end up "stuck" in one corner most of the time. Some aerobatic planes sport an upside-down slip-skid ball on the panel in addition to the normal one. If the G-load is near zero, the ball will become hyper-sensitive (regardless of which way the tube is mounted) -- the least bit of sideforce will send it to the far corner or the tube. A glider's "yaw string" continues to work just fine even when the G-load is zero or negative.
By the way, with a radio-controlled model aircraft, during negatively-loaded (sustained inverted) flight with the aircraft flying away from the operator who is standing on the ground, if the plane is banked away from wings-level inverted, the flight path will curve TOWARD the direction that the operator moved the control stick to initiate change in bank angle.  Again, this is toward the earthward wingtip.
This answer is probably more easily understood if the reader keeps in mind that the direction of the wing's lift vector, and the direction of the G-load, are essentially the same thing (or we could say mirror images of each other.)  If the wing's lift vector points toward the pilot's head, the G-load is positive, and if the wing's lift vector points toward the pilot's feet, the G-load is negative.  If the wing is at the zero-lift angle-of-attack so that there is no lift vector, then the G-load is zero, at least in the up-and-down direction in the pilot's reference frame.

PS a practical approach to the rudder coordination issue in some aircraft, including some rc model aircraft, is to simply avoid making "coordinating" rudder inputs at all during negatively-loaded flight.

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text:
'Gravitational' power vs. engine power
A glider gets the power it needs to fly from the decrease of gravitational potential energy associated to the descent. My question is: for the same weight $W$, airspeed $V$, and a prop efficiency of 100%, is the amount of ‘gravitational power’  $W · w$ that drives a plane in an unpowered glide exactly identical to the ‘engine power’ it needs for s/l, unaccelerated flight?

Related -- https://aviation.stackexchange.com/q/87014/34686

Short answer yes, a powered aircraft at any given airspeed in straight and level flight (note that a glide path is also "level", at constant airspeed) needs to use the same amount of energy to over come drag as the glider.
The potential energy for the glider is altitude. The potential energy for the powered aircraft is fuel.
The conversion of potential energy to kinetic energy gives the aircraft velocity and matches drag forces.
There for, gliding is powered flight! They might "fill up" by catching a thermal, or by having someone tow them up.

The power applied by any given source to a moving vehicle is always equal to the force component exerted by that source in the vehicle's direction of travel, times the velocity of travel.
In steady-state flight, whether we are gliding or flying or horizontally under power, the Drag vector must be opposed by some other force.  In steady-state horizontal powered flight, assuming that the thrust line is exactly parallel to the flight path (as seems to be implied by the question), the Thrust from the motor exactly opposes Drag.  Hence the power applied by the motor equals Thrust * airspeed, which is also equal to Drag * airspeed.
What is the power applied by gravity to an aircraft in gliding flight?
See the three diagrams representing gliding flight illustrated immediately below.  These are three different ways of depicting the same situation-- a steady-state glide at one particular L/D ratio.  On the right-hand diagram, we've included the velocity (airspeed) vector, and its horizontal and vertical components.  The vertical component of the velocity vector is the sink rate. The force vector triangle and the velocity vector triangle are geometrically similar, each having one corner with a 90-degree angle, and one corner equal to the glide angle.

In gliding flight, the net power applied by gravity to the aircraft will be equal to the airspeed times the component of the Weight vector that acts in the direction of the airspeed vector.  The component of the Weight vector that acts in the direction of the airspeed vector is Weight * (cosine glide angle), so we can say that the power applied by gravity is airspeed * Weight * (cosine glide angle).  But airspeed * (cosine glide angle) is the sink rate.  So the power applied by gravity is Weight * sink rate.
Since Drag is the only aerodynamic force that has a component parallel to the airspeed vector-- in fact, the Drag vector is exactly parallel to the airspeed vector-- we can say that all the power applied by gravity to the gliding aircraft is acting to overcome Drag.  It is exactly analogous to the power applied by the motor in level flight.  In gliding flight, the magnitude of the Drag vector is exactly equal to (Weight * cos glide angle).  In other words, in gliding flight, the magnitude of the Drag vector is exactly equal to the component of the Weight vector that acts in the direction of the airspeed vector.
If we could transition from gliding flight to powered flight with no change in airspeed and no change in the magnitude of the Drag vector, then the power required would stay exactly the same.  The source of the power would simply change from gravity to the motor.  The power required in horizontal flight would equal airspeed * Weight * cosine glide angle, which is also equal to Weight * sink rate.
For reasonably high L/D ratios (glide ratios), it's a very good approximation to say that the scenario described above is true.  But in reality, the truth is that when we transition to from gliding flight to powered flight, the Lift vector must become slightly larger.  This means that we must either increase the angle-of-attack and change the lift and drag coefficients, or we must increase the airspeed.  Therefore it is not exactly true to say that the power required in horizontal flight equals Weight times the sink rate that we see in gliding flight at the same airspeed, or at the same angle-of-attack.
Let's explore this some more with some vector diagrams.  To make things clearer, we'll take the case of an aircraft with a very poor L/D ratio of one to one.

The left-hand diagram shows the gliding case.  Note that the Drag vector bears a significant amount of the aircraft weight, which "unloads" the wing.  The Lift vector and the Drag vector each are equal to .71 * Weight.  When we transition to powered flight, we'll hold the angle-of-attack constant, which means that the lift and drag coefficients will stay constant, which means that the Lift/Drag ratio must stay constant.  The only way we can make a closed polygon (in this case a square) out of the Lift, Drag, Weight, and Thrust vectors, while preserving the same 1/1 L/D ratio that we had in gliding flight, is to increase the size of the Lift and Drag vectors, so that the Lift vector (and in this case, also the Drag vector) becomes equal in size to the Weight vector.  Since we haven't changed the angle-of-attack, this can only mean that we've increased the airspeed.  At this particular L/D ratio, the airspeed associated with gliding flight is lower than the airspeed associated with horizontal powered flight by a factor of (square root of .71) = .84.  Both the increased airspeed and the increased Drag vector cause an increase in required power in horizontal flight, compared to the power exerted by gravity in gliding flight at the same angle-of-attack.
In this particular (admittedly extreme) case, the power required for horizontal flight is greater than (weight * sink rate) by a factor of (1/ (.71 * .84)) = 1.68, where "sink rate" means the sink rate in gliding flight at the same angle-of-attack as we have in the powered case.
More generally, the power required for horizontal flight works out to weight * sink rate * / ((cosine ( arctan (D/L)))^1.5), where "sink rate" means the sink rate in gliding flight at the same angle-of-attack as we have in the powered case.
For comparison, here is a table of the increase in power required for horizontal flight beyond the value predicted by (weight * sink rate), for various L/D ratios, where "sink rate" means the sink rate in gliding flight at the same angle-of-attack as we have in the powered case--
L/D 1/1 -- power required for horizontal flight is greater than weight * sink rate by a factor of 1.68
L/D 2/1 -- power required for horizontal flight is greater than weight * sink rate by a factor of 1.18
L/D 5/1 -- power required for horizontal flight is greater than weight * sink rate by a factor of 1.030
L/D 8/1 -- power required for horizontal flight is greater than weight * sink rate by a factor of 1.012
L/D 10/1 -- power required for horizontal flight is greater than weight * sink rate by a factor of 1.0075
Clearly, for most purposes this effect can be considered to be negligible for L/D ratios above 5/1 or so.  But the question did ask whether the power required in the gliding and powered cases was exactly identical.
The original question was phrased to hold airspeed constant between the gliding and powered cases, not angle-of-attack.  To hold the airspeed constant as we transition from the gliding case to the powered case, we must increase the angle-of-attack to increase the lift coefficient to provide the extra lift needed in the powered case.  Now the L/D ratio is almost certainly not remaining constant.  Since airspeed is now constant, the change in the required power will be directly proportional to the resulting change in the Drag vector.  And the resulting change in the Drag vector depends upon where we are on the L/D versus airspeed curve. If we are in high-speed cruise well above the best L/D speed, the Drag vector will actually be smaller in level flight than in gliding flight.  If we are flying slower than best L/D speed, the Drag vector will be larger in level flight than in gliding flight.  There is one special case where we happen to gliding with an angle-of-attack slightly lower than the angle-of-attack yielding the maximum L/D ratio and minimum Drag, and as we transition to horizontal flight, we increase the angle-of-attack and end up with an angle-of-attack slightly higher than the angle-of-attack yielding the maximum L/D ratio and minimum Drag, and the net change of Drag is exactly zero.
Again, these effects will be negligible at reasonably high L/D ratios.  As we transition from the gliding case to the powered case, the Lift vector only needs to increase in magnitude by a factor  of 1 / ((cosine ( arctan (D/L))).  This factor drops to less than 1.01 for L/D ratios higher than 7/1.  The corresponding change in the Drag vector depends on where we are on the L/D curve, but it will clearly be negligible for most practical purposes.
In summary:
for the same weight $W$, airspeed $V$, and a prop efficiency of 100%, is the amount of ‘gravitational power’  $W · w$ that drives a plane in an unpowered glide exactly identical to the ‘engine power’ it needs for s/l, unaccelerated flight?
No, the power requirements are generally not exactly identical in the powered case and the gliding case, because the Lift vector must be slightly larger in the powered case, which implies that the angle-of-attack must be slightly higher in the powered case.  Therefore the Drag vector is unlikely to be identical in size between the two cases.  But for most practical purposes, in aircraft with reasonably high L/D ratios, the difference in Lift, Drag, and power required between the gliding and powered cases is negligible.
See also related answers to related questions:
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- a concise answer with clear diagrams
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- a longer answer that emphasizes many concepts essential to the present answer
Why is the L/D ratio numerically equal to the glide ratio? -- similar to link immediately above
Descending on a given glide slope (e.g. ILS) at a given airspeed— is the size of the lift vector different in headwind versus tailwind? -- the variation in the size of the Lift vector as the still-air glide ratio changes comes into play here, exactly as it does in the present answer
What produces Thrust along the line of flight in a glider?
An airplane has an engine that pushes its flight. What force pushes a glider to fly?
Does lift equal weight in a climb?

Future edit: add missing "W" label to one diagram

This answer is wrong. In any kind of steady-state flight, power = drag * velocity. It is equal in all circumstances for the same drag and velocity. It doesn't matter what's providing the force to balance, whether it's an engine, a tow rope, or gravity.

The answer has been updated to specifically address the last comment, among other issues.

Related -- https://aviation.stackexchange.com/a/87040/34686


text:
Why a kite flying at 1000 feet in "figure-of-eight loops" serves to "multiply the pulling effect of the airflow" on the ship to which it is attached?
In CNN's June 30, 2023 Giant kites could pull cargo ships across the ocean – and slash their carbon emissions

“What differentiates it from other wind solutions,” says Bernatets1, “is that the wing is not just pulled by the wind and countered by the ship.” Instead, it flies in figure-of-eight loops, which multiply the pulling effect of the airflow to give what he calls “crazy power.”
“Plus, we fetch the wind 300 meters above the sea surface, where it’s 50% more powerful,” adds Bernatets. The combination “explains why the power is tremendous for a system that is very compact, simple on the bow of the ship, and can be retrofitted on any ship, not just new ships,” he says.

1Airseas co-founder and CEO Vincent Bernatets
I'm hoping that my question about the aerodynamics and lift from a "giant kite" that "flies in figure-of-eight loops...300 meters above the sea surface" is sufficiently on-topic here.
When kites are used to generate electricity, you certainly need motion - perhaps it reels in and out or it lofts a rotating wind turbine or something else - you need substantial relative motion to generate substantial electrical power.
But a kite affixed to a ship and static relative to it can certainly provide mechanical power to the ship, at least when the wind speed is greater (and in the same general direction) as the ship's speed.
So... what's with the figure-eights?
Question: Why would a kite flying at 1000 feet in "figure-of-eight loops" serve to "multiply the pulling effect of the airflow" on the ship to which it is attached?

different but perhaps somewhat related in that the "magic" happens when two things at different altitudes in media moving in different directions are tethered in a clever way: In which situations can a Dual-Aircraft Platform (atmospheric satellite) remain nearly geostationary?

I've temporarily added both [tag:flow] and [tag:laminar-flow] since "airflow" is mentioned. Once the answer becomes more apparent, tagging can be adjusted. Note that while they are used 18 and 12 times respectively, neither has any usage guidelines currently, and I have a hunch they belong on a lot more questions than that.

A kite that is static relative to the ship is probably stalled, and flying in "8" loops might un-stall the kite (by increasing its forward velocity); if that's true, then it should generate more lift than a stalled kite.

I don't really know either, so not writing an answer. But the kite moving to "fresh" air can have an effect also, because it doesn't build up a region of locally slower flow. Similar to how wind turbine power is relative to the swept area instead of the area of just the blades.

Highly related : https://en.wikipedia.org/wiki/Crosswind_kite_power, https://en.wikipedia.org/wiki/Airborne_wind_turbine.  You get a lot more energy by flying in a dynamic pattern at high speed as opposed to simply maintaining a fixed position relative to the tether point.  Have you ever watched a kite boarder?  Note that the kite is not simply kept in a fixed position relative to the board.

@quietflyer my observations of kite surfers is that the kite is pretty much stationary in relation to the surfer

Consider the apparent wind on the kite, which is the resultant of the actual windspeed (relative to the ship) and the speed of the kite. The fast moving kite sees a much higher apparent windspeed, allowing it to extract more energy. (Similarly, the tip speed of a wind turbine blade is 5-10x the windspeed).

@Jpe61 You haven't observed enough. When kitesurfing you 'park' the kite if you want to cruise, yes.  To do something a more airborne you need more power, which you get this by diving the kite down and back up. Or by looping it, for even more power. In light winds, parking the kite isn't an option and the rider will need to constantly sine the kite up and down just to cruise. Some malfunctions can result in a kite that never stops looping, which is how people get thrown downwind and injured. Bottom line; a moving kite always provides substantially more pull than a parked one.

For sure @aroth, and now that you mentioned it, I must admit that yes, they do at times fly the kite in a pattern, apparently for extra pull. That will quickly become tiresome, so for longer cruises they will leave the kite stationary.

A racing sailboat driven only by such a kite is undergoing sea trials.  Many delicious technical details: https://www.yachtingworld.com/extraordinary-boats/armorkite-650-kite-surfing-yacht-sailing-126194

@CamilleGoudeseune oh that is so cool! "The ArmorKite is extremely sensitive on the tiller, and also sensitive to the positioning of the crew, whose total weight can easily equal that of the boat itself. It’s important to keep a close eye on both longitudinal and lateral trim." I don't know what much of that means, but I get the feeling that if the point of attachment could be moved dynamically somehow (slides along a front-to-back rail) rather remain at a fixed point...

@CamilleGoudeseune ...they might perhaps add even more control. I'm going to go buy a dual-line kite now, stand firmly on the shore and away from everyone else, and get in on the fun!

The giant kites mentioned aren't the quadrilateral kites kids fly. They are airfoil shaped parachutes with open noses that generate lift as they move through the air. Ram air pressure inflates them and maintains their shape. In skydiving lift works against gravity. (ignoring turns and the like)

Anchored to bow of a boat the airfoils must travel perpendicularly to the direction of ship motion in order to pull it forward. Here lift works against hull drag in the ocean. The farther the parachute gets from perpendicular, the less forward force it provides so it has to turn around to make another pass across the bow. The path that allows the wing to maintain continuous flight while generating the maximum forward pull is a horizontal figure eight. There is nothing magic about the shape of the path. Altitude is selected for best wind and size of parachute.
An airfoil generates more pull than a simple round parachute of the same area placed directly in front of the ship because it is flying at speed, one reason being the higher dynamic pressure experienced because of that speed. Looking at the weight each type can carry per unit area is one way to see this difference, recognizing that each type has a range of performance capability. A round T10 parachute with an inflated diameter of 25.7ft has an area of 480sqft and carries a maximum load of 350lb, or 0.72 lb/sqft. The ram air RA1 parachute has an area of 360sqft and carries a maximum load of 450lb, or 1.25 lb/sqft. In addition, the round parachute has a descent rate of 22 fps while the ram air chute's descent rate is 8 fps, so the wing will work in much lighter wind.

Possibly interesting for everyone here: https://www.diva-portal.org/smash/get/diva2:1270437/FULLTEXT01.pdf (esp. Fig. 2 on p. 9! So they could EASILY add a swash plate, and a much simpler one than what I suggested above!) and also https://mdpi-res.com/d_attachment/energies/energies-16-02603/article_deploy/energies-16-02603-v2.pdf

Re "Anchored to bow of a boat the airfoils must travel perpendicularly to the direction of ship motion in order to pull it forward." -- Suggested edit for clarification-- something along the lines of "An airfoil is designed to generate lift with minimal drag. To pull the boat using lift rather than drag, the airfoil must travel perpendicular to the direction of the relative wind experienced by the ship." Pertains to this related question: https://aviation.stackexchange.com/q/99840/34686. Technically, the kite could pull the boat using drag alone... remaining directly downwind of the boat--

Or for more detail see  https://aviation.stackexchange.com/questions/99840/forces-acting-on-ram-air-parachute-for-pulling-ships#comment278607_99840 and following comment--

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Have you ever flown a dual line kite? If you keep it steady it will climb until it reaches a steady state and until it is mostly horizontal and only pull minimally on the lines.
If however you turn it and have it go near the ground the kite will be more vertical into the wind and it will pull a lot more on the lines. And the entire goal of flying a kite on a ship is to get it to pull on the lines.
You could make the kite go in circles or ovals but that will end up tangling the lines. So a figure 8 is used instead.

"Have you ever flown a dual line kite?" It's been quite a while - and these days it's kind-of dangerous here But maybe I'll get (a smaller) one and give it a try, or just watch someone (from a safe distance)

I think  the reason is to maintain the optimum angle of attack, while also positioning the kite lines so they pull the ship forward. It probably only helps when the ship is sailing at certain angles to the wind.
Wings generally work best when at 10-15 degrees to the airflow. That probably works well for a ship sailing across the wind. The kite can be ahead of the ship, and the lines pull in the right direction.
However, when the ship is sailing downwind, putting the kite in front of the ship would leave it at 90 degrees to the wind, and deeply stalled. Moving the kite to the side would allow it fly efficiently, but then the pull wouldn't be in a useful direction. Flying figure-8s in front of the ship allows the kite's airspeed to keep the angle of attack below the stall, while keeping the lines pulling mostly forward.
Additionally, the faster the kite is flying, the more lift it generates. If the kite has a lift:drag ratio of 5:1, it could fly 5 times faster than windspeed. Lift is proportional to speed squared, so it could generate 25 times as much force than if it were stationary relative to the ship. When you fly a steerable kite, it's very noticeable how much more it pulls when moving fast!

Think about it in terms of how much air your sail is interacting with:

The classical sail of a ship with the wind at its back can only interact with the air directly around the ship. The ship's speed is subtracted from the wind speed when computing the effective air speed of the sail, which is lower than the wind speed by consequence. That's not much of a reaction mass that the sail interacts with per second, and hence not a lot of force can be generated. As such, sailing in front of the wind is always rather slow.

A sailing boat that's sailing at a right angle with the wind adds its own speed to the wind speed (vector addition). This allows the effective wind speed around the sail to get much higher, allowing the sail to interact with a lot more air per second. And, the faster the boat, the greater the force the sail can generate!

An airfoil shaped parachute that's moving perpendicular to the wind is not restricted to the speed of the ship. As such, it can move even faster, dominating the relative wind speed with its own speed. Again, this vastly increases the amount of air it can interact with, allowing it to pull much, much harder on its line than any fixed sail could.

Flying the kite in figure 8 serves to keep the kite's speed as high as possible, and by consequence its effective wind speed and reaction mass per second to interact with.
The same concept of maximizing air reaction mass applies everywhere in aviation. It's the same reason why helicopters need to work harder when hovering than when they are in straight flight, and why air planes are generally more efficient the faster they fly. Because more reaction mass means that the reaction mass does not need to be accelerated to as high speeds to provide the same force, and thus that less energy is carried away by the accelerated reaction mass.

Of course, they could also try this, or this, but these are a bit faster.  These have "crazy power", able to go  many times faster than the wind.  Reduction of drag (on ice) definitely helps.
But these make most efficient use of the wind for lifting force.
The kite flying "figure 8s" is combining its motion and the direction of the wind to create a relative wind most favorable for the creation of lift.  Another way of doing this is to sail the entire ship around 45 degrees off of downwind back and forth with a fixed sail on a mast. This is what iceboats can do, enabling them to reach speeds several times more than the wind.
The kite technique allows the ship to stay on course, and harnessing the stronger winds above the surface of the ocean provide very promising advantages.

Link-only answer‽

These five links are only comments.  They could be omitted without affecting the meat of this answer.

Ah, but one does not live on meat alone.  Just adding some foundation to the concept, and I'm not only supporting it, but considering to suggest an even larger "air sail" supported by blimps.


text:
Why are clearing turns supposed to be shallow bank?
During my CFI training, my instructor relayed a story in which he had a mid-air collision while teaching (how to teach) clearing turns to two other CFI applicants. The other aircraft was underneath them and may have climbed into them. No one in his aircraft saw the other aircraft. They did not even know they had hit another aircraft until they heard the report of another crash nearby.
My question is (in the title) why is a shallow banked clearing turn considered better than a steep one? I feel like I get a much better view of everything (above, level, and below) in the direction of the turn if it is a steeper turn. The downside is that I'm changing direction more quickly...but I think seeing other aircraft wins over that. Furthermore, large, rocking wing movements tend to make the aircraft I am in easier to see as opposed to gentle, shallow turns.

Are you in a high wing or low wing aircraft?

High wing. I can see where the argument for a shallow turn came from. Thank you.

Clearing turns solve two problems: seeing other aircraft and helping them to see you.
Banking will make you more visible to other aircraft, who may not have noticed you before. A lower bank angle (and thus turn rate) gives them more time to see and avoid you. But a very low bank angle also doesn’t make you as visible in the first place, so it’s a balance.
I was taught 20-30°, and I’ve never had anyone say later that was too much or too little. I also don’t have to worry much about losing altitude at moderate bank, so I can keep my eyes outside scanning for traffic—not checking the altimeter every few seconds like during “real” maneuvers.

Generally because one is actively looking out the window for other aircraft, not at the instruments or the horizon.  This is a distraction that places the less experienced pilot (doing a clearing turn before practicing a maneuver) farther away from the safety of "straight and level".  Much better to mistakenly turn a shallow turn into a steeper turn than to turn a steeper turn into an extreme bank.
However, too shallow a bank takes more time, and taking too long increases the chance of another plane entering the practice area undetected.
With two people (pilot and instructor), the turn can be fairly aggressive (more than 30 degrees), itself a practice maneuver, but an inexperienced pilot alone could quickly get into trouble by overbanking.
A good way to do a clearing turn is to slow down below cruise speed, which reduces the turn radius:

r = V$^2$/g x tangent(bank angle)

but we must be aware that the bank angle increases stall speed by:

turning stall speed = stall speed x square root(load factor)

Load factor in turn is calculated as:

1/cosine(bank angle)

That is a matter of personal preference. The FAA's Airplane Flying Handbook only states: "Pilots should always clear the area with two 90° clearing turns looking to the left and the right, as well as above and below the airplane." This article from Flying Magazine, "Clearing Turn Basics", also does not mention bank angles in clearing turns. Personally, I make shallow clearing turns, and then decide if I want to practice steep turns. Someone flying a biplane like a Pitts Special, for example, might have a different opinion.


text:
Longitudinal/Lateral CG envelope for hoist operations
Does a Rotorcraft hoisting CG chart show what the CG will be as the load is hoisted from top to bottom or vice versa?

Can you give an example of a "Rotorcraft hoisting CG chart", or a link to one? Is this just a hypothetical thing that you are imaging might exist, or something that you've actually seen?  (See last para of my answer as to why such a thing might exist.)

The dynamics of an aircraft with something dangling from a single long line have little in common with an aircraft with the same object attached below the aircraft by a completely rigid (long) rod or pylon.
So we won't find any chart that treats raising or lowering an object with a hoist, as being equivalent to a vertical shift in a helicopter's CG.
Incidentally, that's why hang gliders, where the pilot suspension system can be modeled as a single flexible line, at least when the pilot is not using his or her muscles to make a control input, don't have the same hands-free pitch and roll stability as do paragliders, where the multiple suspension lines keep the pilot essentially rigidly fixed in position relative to (and well below) the rest of the aircraft.
In the hang glider case, when the pilot exerts no muscle force with his arms, or releases the control bar completely, he is free to swing fore and aft and side to side and so his weight acts as if it is located at the point where the flexible tether (called a "hang strap") connects to the rigid structure of the glider.  Not as if he is rigidly fixed in position below the glider.
The same will be true during helicopter hoisting operations, at least as long as the hoisting rate is constant, and assuming that the object is not "penduluming" or swinging wildly, which can change the object's apparent weight due to centrifugal loading.  If the object is simply dangling at the end of a line which is fixed or being raised at lowered at a constant rate, for stability purposes the object can be viewed as being located at the upper end of the hoisting line, i.e. at the point where the hoist line meets the pulley of the winch.
For a given hoist location, one might well imagine that only a certain part of the total allowable helicopter CG envelope might actually be compatible with hoisting operations, and that this might vary according to the weight of the object being hoisted.  Is this perhaps what you've seen a chart relating to?  If so, the length of the hoist line would not matter, so nothing changes as the object is raised or lowered, due to the dynamics explained above.

Presuming that the winch is in a fixed location, (not on a track that allows it to move fore and aft or sideways) the longitudinal and lateral center of gravity of the helicopter would not change as the load is raised or lowered.


text:
Why doesn’t the minimum drag point meet the point at the best Cl/CD polar?

On the first picture, it shows that the minimum drag point is exactly the same as the maximum L/D ratio where it also supposes to be at The point as it shows on the second picture. However, according to the third picture, it doesn’t look like the minimum drag point is at the maximum L/D on the second picture. Why does this happen?

simple answer: because lift is not constant and also increases with aoa

You need to distinguish between coefficients and the forces.
Coefficients are dimensionless numbers which are used to compare flow phenomena at different speeds, scales or atmospheric conditions. The force coefficients are the forces normalised by a reference area (to remove size effects), speed squared divided by two (to remove speed effects) and air density (to remove atmospheric effects). Mathematically speaking: $$c_X = \frac{X}{\rho\cdot\frac{v^2}{2}\cdot S_{ref}}$$ with $X$ being the force, $\rho$ being density, $v$ being air speed and $S_{ref}$ the reference area.
The lowest drag coefficient is at rather a low lift coefficient. In order to create enough lift to carry its own weight, the aircraft has to fly rather fast there. This high dynamic pressure at this polar point means that the total drag force is higher than at a polar point slightly above it.
At the point of best L/D the ratio between both forces reaches a minimum, so actually the lowest drag for a given lift is at this point.
Only if you disregard lift (say, in a vertical dive where the goal is to reach the highest speed) will the point of the lowest drag coefficient also be the point of lowest drag (at the same dynamic pressure). To generalize: The minimum drag point shifts between the maximum L/D polar point and the minimum $c_D$ polar point with the cosine of the flight path angle. Between 90° and 270° flight path angle (inverted flight) it shifts between the minimum $c_D$ point and the minimum L/D polar point (where lift is negative).

Sorry, I couldn’t get you. Do you mean if it is a level flight, then the minimum drag  point (only for level flight) is also the maximum L/D and CL/CD while in any other condition, the minimum drag point is at the point of minimum CD which was decided when the aero foil was made?

I guess the 4.3 picture makes this confusing when it says ‘Minimum Drag’ and points at the blue 2deg. It should say : Minimum Drag Coeficient. Actual drag Force at 2deg is higher than at 6 deg ( for a given unaccelerated flight)

@RobertLo: Yes in case of level flight, but for climb and descent it is still very near the maximum L/D point since sufficient lift needs to be created. Only in the vertical dive is no lift required, and only then will minimum drag be where minimum $c_D$ is. The minimum drag point shifts between optimum L/D and minimum $c_D$ with the cosine of the flight path angle.

"Minimum drag" at 2deg in image 4.3 is just minimum Cd (coefficient not force). It is the most streamlined position and you will generate the least amount of drag here, for a given speed. You won't generate much lift though, if any. In fact, if you want to fly level (ie. at 1g) you will have to fly so fast that the drag will still become expensively large. Larger that if you were to fly at a more optimum angle of 5deg.
If Cl was constant over aoa, the best Cl/Cd would indeed occur at minimum Cd. ( a fixed quantity divided by the smallest ammount), alas Cl and Cd both increase with increasing aoa, reaching a best Cl/Cd ratio (in your example) at 5deg in image 4.7.
After that point, the Cd will increase a lot more than Cl, thus making the ratio worse.

What I confuse is that on the third picture,why is the minimum drag point not at maximum L/D and CL/CD?

forget that this is about aerodynamics, just focus on the graph: on any polar graph like the left image 4.3 , the highest ratio Y / X value will occur on the tangent from origin (ie 5 deg in your image, see 4.7) not on the apex of MinX ( at 2 deg in your image).This is true for any polar graph; It’s math, not aerodynamics

Re "Minimum drag is just that, minimum Cd." -- actually, not.  For a different view see https://aviation.stackexchange.com/a/86327/34686.

Yeah, should have put that in quotes as I was referring to the label on the graph. I tried my best at rephrasing it

On the first picture, it shows that the minimum drag point is exactly
the same as the maximum L/D ratio

This makes sense, at least in the context of linear constant-altitude flight, because Lift must equal Weight and therefore is constant, so if we maximize L/D, we also must minimize Drag.  (Also, if we're maximizing L/D, we're simultaneously maximizing Cl/Cd.)
But since airspeed is not fixed as angle-of-attack varies, it does not follow that by minimizing Drag, we are also minimizing Cd-- and that is the key to solving your quandry.
Despite the misleading label on the third graph, the flagged point is actually not the point of minimum Drag.  Rather, it is the point where the drag coefficient is minimized.  In the context of linear constant-altitude flight, where L = W but airspeed is free to vary as angle-of-attack varies, the point of minimum drag coefficient is completely different from the point of minimum Drag.

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